I was reading this question link text
and can't seem to see why, if $\pi: P \to B$ is a principle $G$-bundle and $$\rho:G \to GL_n(\mathbb{C})$$ is a representation of $G$, then the total space $P \times_{\rho} \mathbb{C}^n$ is locally trivial.

You're making $P \times_\rho \mathbb C^n$ the bundle over $B$ with fiber over a point $b \in B$ equal to $\pi^{-1}(b) \times_G \mathbb C^n$, right, meaning you're modding out by the diagonal action of $G$. So in effect you're `replacing' the $G$ fibers of $\pi$ by $\mathbb C^n$ fibers.
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Ryan BudneyFeb 2 '10 at 19:11

This is standard differential topology/geometry and can be found in any undergraduate texts on this subject. I suggest you look in one of those - it's probably in Hatcher's book (available online), though I don't know for sure. You'll also gain a better understanding of the topic by reading around this result in the books than the answers you are likely to get here will give you.
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Loop SpaceFeb 2 '10 at 20:54

Really? Gosh! I'll have to look that up next time I'm near an undergraduate diff geom/top book. That's shocking, if true! Ah, well. I'm pretty sure it's in Kriegl and Michor so I could always refer to that.
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Loop SpaceFeb 2 '10 at 21:14

4 Answers
4

A principal G-bundle $\pi: P \to B$ is locally equivalent to a product. Depending on who you ask, this is either part of the definition, or a short lemma. It means that there is a cover of B by open sets U, together with bundle isomorphisms $\alpha_U: \pi^{-1}U \to G \times U$ that are both G-equivariant, and induce transition sections in G: see Wikipedia.

The associated bundle construction is a quotient of $P \times \mathbb{C}^n$ by an equivalence relation given by the actions of G on the left and right factors. Since $\alpha_U$ is G-equivariant, the map $\alpha_u \times id: \pi^{-1}U \times \mathbb{C}^n \to G \times U \times \mathbb{C}^n$ is also a G-equivariant bundle map, so you get bundle isomorphisms on the quotient bundles: $\alpha_u \times_\rho id: \pi^{-1}U \times_\rho \mathbb{C}^n \to U \times \mathbb{C}^n$.

You can trivialize that bundle over the same open sets on which $\pi$ is trivial with essentially the same trivialization---up to composing with $\rho$.

Later: The easiest way to see this is, I think the following. If your initial principal $G$-bundle is trivial, then it is more or less evident that $P\times_\rho\mathbb C^n$ is also trivial. Now if $\pi$ is not trivial but its restriction $\pi|\_U:\pi^{-1}(U)\to U$ is trivial over an open subset $U\subseteq B$, then by the previous observation and a little unravelling of the notation, we have that $P\times_\rho\mathbb C^n$ is trivial over $U$.

@Scott: I'm going to add this as an answer because it a bit long for a comment: Sorry, but I'm a bit slow and want to be certain that I understand exactly what you're saying: If $(p,v) \in P \times_{\rho} \mathbb{C}^n$, then $$(\alpha_u \times id)(p,v) = (\pi(p),g_p,v) \in U \times G \times \mathbb{C}^n,$$ for some $g_p \in G$, depending on the choice of $\alpha_U$. This is not in $U \times \mathbb{C}^n$ as we would want, moreover it's not even well defined. So by $\alpha_U \times_{\rho} id$ do you mean the composition
$$
(id \times m) \circ (id \times \rho \times id)\circ (\alpha_U \times id),
$$
that maps
$$
(p,v) \mapsto (\pi(p), \rho(g_p)v)?
$$

I've never worked this out carefully, but in my usual backwards "I prefer vector bundles" approach, I would do the following: First, assume that the representation of $G$ is faithful. Then you can view any local trivialization of the principal bundle as giving all possible $G$-frames of a trivialization of a vector bundle (i.e., the columns of each matrix representing an element in $G$ form a basis of the fiber of the vector bundle). Using this perspective, you can infer from the gluing maps for the principal bundle how to glue together the trivializations of the vector bundle.